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A certain combinatorics problem requires finding the determinant of matrices like this:

\begin{equation} \begin{bmatrix} a & b & c & d & f \\ a^2 & b^2 & c^2 & d^2 & f^2 \\ a^m & b^m & c^m & d^m & f^m \\ a^{m+1} & b^{m+1} & c^{m+1} & d^{m+1} & f^{m+1} \\ a^{m+2} & b^{m+2} & c^{m+2} & d^{m+2} & f^{m+2} \end{bmatrix} \end{equation}

For each column, is an arithmetic progression of powers and then another arithmetic progression for ${m+y}$ for integer $m$

What is this type of matrix called? I cannot find any information about matrices of this form or a book or paper that describes how to find determinants of this type of matrix.

In the above example, the has $C(5,3)=10$ permutations of triples of $a^m,b^m,c^m,d^m,f^m $ and then polynomials of $a,b,c,d,f$ associated with each in order to construct the determinant. One is $(a-b)(c-d)(c-f)(d-f)(cdf)^{m}$

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    $\begingroup$ The ratio between the determinants that you want and the classical Vandermonde determinant is measured by a certain symmetric polynomial called a Schur Polynomial. These have all sorts of combinatorial expansions/representations that might be useful depending on your context. $\endgroup$
    – Gjergji Zaimi
    Commented Jul 19, 2022 at 0:03
  • $\begingroup$ Perhaps their most important interpretation is as the trace of a matrix in $GL_n$ (for $n=5$ in this case) acting on an irreducible representation. $\endgroup$
    – Will Sawin
    Commented Jul 19, 2022 at 0:29
  • $\begingroup$ math.stackexchange.com/questions/3300187/… this question is related, the last column is a plus 1 exponent... $\endgroup$
    – Toni Mhax
    Commented Jul 20, 2022 at 9:20

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It does not have an elegant solution

Too many partitions. The reason why the original vandermonde matrix has a nice form is because there is one partition regardless of the size.

So for the 7x7 version of the above matrix, we have 4 parts which have powers of m and 3 which do not. So the determinant will be of the form in which each 4-permutation of the 7 variables, for 35 different products.

Consider that $a,b,c \approx z^{1/3}$ and $d,f,g,h \approx z$ as expansion of infinite series

So the determinant will be of the form $f_1(z)z^{4m}+f_2(z)z^{10m/3}+f_3(z)z^{8m/3}+f_4(z)z^{2m}$

So we have $C(4,1)C(3,3)+C(4,4)+C(4,3)C(3,1)+C(3,2)C(4,2)=C(7,4)=35$

$f_1(z)z^{4m}$ is the easiest and only has a single permutation, that being $C(4,4)$. This means that the $a^{m+y},b^{m+y},c^{m+y}$ terms can be zeroed-out, so you have a partition block matrix with one of the blocks zero, and determinant is easy to evaluate on this being the product of two small regular vandermonde matrices determinants, which is the product of 9 pairs (three for the 3x3 matrix and six for the 4x4 one). This can allow one to determine the asymptotic behavior of such matrices.

The next one has 12 permutations, so the determinant gets really messy fast. This require zero-ing out triples of a,b...h and then swapping rows to make additional block matrices.

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